Answer
$y''=\frac{1}{e^2}$
Work Step by Step
Remember that $e$ is simply a constant value.
To find what $y$ is when $x=0$, plug it into the equation.
$(0)y+e^y=e$
$e^y=e$
$y=1$
So we now know our $(x,y)$ coordinate is $(0,1)$.
Now we use implicit differentiation to find the first and second derivatives.
$(xy+e^y=e)'$
(1) $xy'+y+e^y\dot\thinspace y'=0$
Solve for $y'$
$y'=\frac{-y}{x+e^y}$
$y'=\frac{-1}{0+e^1}=\frac{-1}{e}$
From equation (1):
$(xy'+y+e^y\dot\thinspace y'=0)'$
(2) $xy''+y'+y'+e^y\dot\thinspace y'y'+e^y\dot\thinspace y''=0$
Solve for $y''$ and plug in $y'=\frac{-1}{e}$, $x=0$, and $y=1$.
$y''=\frac{-2y'-e^y\dot\thinspace y'y'}{x+e^y}$
$y''=\frac{1}{e^2}$
